8 puzzle Problem using Branch and Bound in C - GeeksforGeeks (2024)

// C Program to print path from root node to destination// node for N*N -1 puzzle algorithm using Branch and Bound// The solution assumes that instance of puzzle is solvable#include <limits.h>#include <stdio.h>#include <stdlib.h>#include <string.h>#define N 3// state space tree nodesstruct Node { // stores the parent node of the current node // helps in tracing path when the answer is found struct Node* parent; // stores matrix int mat[N][N]; // stores blank tile coordinates int x, y; // stores the number of misplaced tiles int cost; // stores the number of moves so far int level;};// Function to print N x N matrixvoid printMatrix(int mat[N][N]){ for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) printf("%d ", mat[i][j]); printf("\n"); }}// Function to allocate a new nodestruct Node* newNode(int mat[N][N], int x, int y, int newX, int newY, int level, struct Node* parent){ struct Node* node = (struct Node*)malloc(sizeof(struct Node)); // set pointer for path to root node->parent = parent; // copy data from parent node to current node memcpy(node->mat, mat, sizeof(node->mat)); // move tile by 1 position int temp = node->mat[x][y]; node->mat[x][y] = node->mat[newX][newY]; node->mat[newX][newY] = temp; // set number of misplaced tiles node->cost = INT_MAX; // set number of moves so far node->level = level; // update new blank tile coordinates node->x = newX; node->y = newY; return node;}// bottom, left, top, rightint row[] = { 1, 0, -1, 0 };int col[] = { 0, -1, 0, 1 };// Function to calculate the number of misplaced tiles// i.e. number of non-blank tiles not in their goal positionint calculateCost(int initial[N][N], int final[N][N]){ int count = 0; for (int i = 0; i < N; i++) for (int j = 0; j < N; j++) if (initial[i][j] && initial[i][j] != final[i][j]) count++; return count;}// Function to check if (x, y) is a valid matrix coordinateint isSafe(int x, int y){ return (x >= 0 && x < N && y >= 0 && y < N);}// print path from root node to destination nodevoid printPath(struct Node* root){ if (root == NULL) return; printPath(root->parent); printMatrix(root->mat); printf("\n");}// Comparison function to be used to order the heapint comp(const void* lhs, const void* rhs){ struct Node* l = *(struct Node**)lhs; struct Node* r = *(struct Node**)rhs; return (l->cost + l->level) - (r->cost + r->level);}// Function to solve N*N - 1 puzzle algorithm using// Branch and Bound. x and y are blank tile coordinates// in initial statevoid solve(int initial[N][N], int x, int y, int final[N][N]){ // Create a priority queue to store live nodes of search // tree struct Node* pq[100000]; // Priority queue with sufficient size int pqSize = 0; // create a root node and calculate its cost struct Node* root = newNode(initial, x, y, x, y, 0, NULL); root->cost = calculateCost(initial, final); // Add root to list of live nodes pq[pqSize++] = root; // Finds a live node with least cost, // add its children to list of live nodes and // finally deletes it from the list. while (pqSize > 0) { // Sort priority queue based on cost + level qsort(pq, pqSize, sizeof(struct Node*), comp); // Find a live node with least estimated cost struct Node* min = pq[0]; // The found node is deleted from the list of live // nodes for (int i = 1; i < pqSize; i++) pq[i - 1] = pq[i]; pqSize--; // if min is an answer node if (min->cost == 0) { // print the path from root to destination printPath(min); return; } // do for each child of min // max 4 children for a node for (int i = 0; i < 4; i++) { if (isSafe(min->x + row[i], min->y + col[i])) { // create a child node and calculate its // cost struct Node* child = newNode( min->mat, min->x, min->y, min->x + row[i], min->y + col[i], min->level + 1, min); child->cost = calculateCost(child->mat, final); // Add child to list of live nodes pq[pqSize++] = child; } } }}int main(){ // Initial configuration // Value 0 is used for empty space int initial[N][N] = { { 1, 2, 3 }, { 5, 6, 0 }, { 7, 8, 4 } }; // Solvable Final configuration // Value 0 is used for empty space int final[N][N] = { { 1, 2, 3 }, { 5, 8, 6 }, { 0, 7, 4 } }; // Blank tile coordinates in initial configuration int x = 1, y = 2; solve(initial, x, y, final); return 0;}